What are the solutions to the equation >0=x^2-1 ?

The solutions to the equation >0=x^2-1 are x=sqrt(>0+1),x=-sqrt(>0+1)

Find the zeros of >0=x^2-1

The zeros of >0=x^2-1 are x=sqrt(>0+1),x=-sqrt(>0+1)

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> 0 = x^{2} - 1

x = > 0 + 1, x = - > 0 + 1

Solution steps

> 0 = x^{2} - 1

Switch sides

x^{2} - 1 => 0

Move

1

to the right side

x^{2} => 0 + 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = > 0 + 1, x = - > 0 + 1

9 x^{2} + 54 x = 0

81 a^{36} - 64 b^{16}

5 n^{6} + 2 n^{5}

- 6 (1 + 6 x) < 6 (1 - 5 x)

\frac{\frac{1}{a} + \frac{1}{a + 1}}{\frac{1}{a} - \frac{1}{a + 1}}

What are the solutions to the equation >0=x^2-1 ?
The solutions to the equation >0=x^2-1 are x=sqrt(>0+1),x=-sqrt(>0+1)
Find the zeros of >0=x^2-1
The zeros of >0=x^2-1 are x=sqrt(>0+1),x=-sqrt(>0+1)